Question: You have found the following ages (in years) of all $5$ zebras at your local zoo: $ 8,\enspace 11,\enspace 17,\enspace 7,\enspace 19$ What is the average age of the zebras at your zoo? What is the standard deviation? Round your answers to the nearest tenth. Average age: $ $
Solution: Because we have data for all $5$ zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$. To find the population mean, add up the values of all $5$ ages and divide by $5$. $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{8 + 11 + 17 + 7 + 19}{{5}} = {12.4\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $8$ years $-4.4$ years $19.36$ years $^2$ $11$ years $-1.4$ years $1.96$ years $^2$ $17$ years $4.6$ years $21.16$ years $^2$ $7$ years $-5.4$ years $29.16$ years $^2$ $19$ years $6.6$ years $43.56$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean, we can find the variance $({\sigma^2})$, without introducing any bias, by simply averaging the squared deviations from the mean: $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{19.36} + {1.96} + {21.16} + {29.16} + {43.56}} {{5}} $ $ {\sigma^2} = \dfrac{{115.2}}{{5}} = {23.04\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$. ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{23.04\text{ years}^2}} = {4.8\text{ years}} $ The average zebra at the zoo is $12.4$ years old. There is a standard deviation of $4.8$ years.